Let $\Omega =B(0)=\{z\in\mathbb{C}\cong\mathbb{R}^2 : \lvert z\rvert<1\}, y=0,$ and define
$h(t,z)=\begin{cases} \lvert z\rvert, & t=0, z\in\overline\Omega \\ \lvert z\rvert \exp(i\phi/t), & 0<t\leq1, z=\lvert z\rvert e^{i\phi}, 0\leq\phi\leq2\pi t \\ \lvert z\rvert, & 0<t<1, z=\lvert z\rvert e^{i\phi}, 2\pi t<\phi\leq2\pi \end{cases}$
We have that $h(t,\cdot)$ and $h(\cdot,z)$ are continuous. I'm trying to show that h is not continuous as a function $\mathbb{R}^3\rightarrow\mathbb{R}^2$. I have trouble thinking of a correct sequence of points to obtain a contradiction with Heine's definition of continuity.
Let $t_n=1/n$ and $z_n = e^{i\pi/n} = e^{i\theta_n}$, where $\theta_n = \pi/n$. Then $(t_n,z_n)$ converges to $(0,1)$ and $h(0,1)=1$. But $2\pi t_n = 2\pi/n > \pi/n=\theta_n$, so $h(z_n,t_n)=e^{i\theta_n/t_n} = e^{i\pi} = -1$, thus $h(z_n,t_n)\not\to h(0,1)$, and then $h$ is not continuous.